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From: George Levy <GLevy.domain.name.hidden>

Date: Sun, 08 Apr 2001 13:52:45 -0700

Marchal wrote:

*> George Levy wrote:
*

*>
*

*> >My binder is getting fatter but I am catching up...
*

*>
*

but it is getting fatter....

I feel like Zeno... Or maybe like Sisyphus....:-)

*>
*

*> You are welcome.
*

Thanks

*> Theorems:
*

*>
*

*> 1. (W,R) respects []p -> p iff (W,R) is reflexive,
*

*>
*

*> 2. (W,R) respects []p -> [][]p iff (W,R) is transitive,
*

*>
*

*> 3. (W,R) respects p -> []<>p iff (W,R) is symmetric,
*

*>
*

*> 4. All frames respect [](p->q) -> ([]p -> []q).
*

*>
*

I am not sure if you gave a definition of reflexive, transitive and symmetric

*>
*

*> Let us prove 1. and 4. (I leave 2. and 3. as exercice)
*

*>
*

I find it easier to comprehend these theorems if I state them in English:

1) If p is true in all worlds in set R, then p is true in world A. This is true

if A belongs to set R. (we could use a Venn diagram to prove this) ...... OK

2) if p is true in all worlds in set T then p is true in all worlds in set T in

all worlds in set T (????) There seems to be two levels of "in all worlds"

Slightly confusing

I don't get this one.... in English.

3) If p is true in world A then in all worlds in set R there is at least one

world B where p is true. This statement is true if A =B........ OK

4) If the relation "p implies q" is true in all worlds in set S, then, if p is

true in all world in S then q is true in all world in S... OK

*>
*

*> Proof of 1.
*

*>
*

*> Because of the "iff" we must prove two things.
*

*>
*

*> we must prove a) if R is reflexive then (W,R) respects []p -> p,
*

*> and we must prove b) if (W,R) respects []p -> p then R is reflexive.
*

*>
*

*> Let us prove a).
*

*> To prove that "if A then B", a traditional way consists in supposing
*

*> that we have A and not B, and showing that this leads to a
*

*> contradiction. (BTW you can verify that ((A & -B) -> FALSE) <-> (A->B))
*

*> is a tautology).
*

*> So, let us suppose that R is reflexive and that (W,R) does not
*

*> respect []p -> p.
*

*> To say that (W,R) does not respect []p -> p means that there is at
*

*> least one world w in which []p -> p is false. Remember that worlds
*

*> obeys classical logic, so that if []p -> p is false at w, it means
*

*> that []p is true at w and p is false at w (and -p is true at w).
*

*> But (by Kripke semantics) if []p is true at w, it means p is true
*

*> at every world accessible from w. But R is reflexive, so we have wRw.
*

*> So p and -p are true at w. But then w does not obey to classical logic.
*

*> Contradiction.
*

*>
*

*> Let us prove b), that is: if (W,R) respects []p -> p then R is
*

*> reflexive.
*

*> Suppose (W,R) respects []p -> p and that R is not reflexive.
*

*> And let us search a contradiction from that.
*

*> If R is not reflexive, it means that there is a world w in W such that
*

*> we don't have wRw. Let us build a model on that frame by defining
*

*> a valuation V such that V(p) = FALSE in w, and V(p) = TRUE on all
*

*> world accessible from w (if there is any(°)).
*

*> By Kripke semantics we have []p true in w. So now we have both
*

*> []p and -p true at w. But we have suppose that (W,R) respects
*

*> []p -> p, so []p -> p is true at w, so p is true at w (because []p
*

*> has been shown true at w, and w obeys classical logic). But then
*

*> again p and -p are true at w. Contradiction.
*

*>
*

*> OK? For theorems 2. and 3. you should perhaps find the proofs
*

*> yourself. It is almost more easy to find the proof than to read
*

*> someone else proof. The only difficulty is in the case b) and
*

*> consists in finding the valuation which leads to the contradiction.
*

*> Hint: put V(p) = FALSE in w, and V(p) = TRUE on all
*

*> world accessible from w; or, for 3. put V(p) = TRUE in w, and
*

*> V(p) = FALSE on all world accessible from w.
*

*>
*

*> (°) Hey! but what if w is a cul-de-sac or terminal world?
*

*> Then []p, or even []False is always true in w. This follows
*

*> easily from classical logic: []p is true at w because
*

*> "(it exists x such that wRx) entails p true at x", this
*

*> last sentence is true because wRx is always false for a
*

*> terminal world, and "false implies whatever you want" is
*

*> always true.
*

*> Another way to swallow this tricky point is to remember
*

*> that []p is -<>-p, but -<>(what you want) must be
*

*> automatically true in a terminal world, by simple
*

*> Kripke semantics for <>.
*

*> Terminal world are are disquieting: everything is necessary
*

*> and nothing is possible in a terminal world.
*

*>
*

*>
*

*> Proof of 4.
*

*> We want to show that [](p->q) -> ([]p -> []q) is respected in all
*

*> frames. This means [](p->q) -> ([]p -> []q) is true in all world
*

*> in all model in all frame!
*

*>
*

*> It will be easier to show that the theorem holds for the equivalent
*

*> sentence ([](p->q) & []p) -> []q
*

*> Do you see the equivalence? It is just propositional logic:
*

*>
*

*> A -> (B -> C) <-> (A & B) -> C
*

*>
*

*> Suppose now that ([](p->q) & []p) is true in some world w.
*

*> We must show []q is true in w.
*

*> If w is a terminal world then []q is automatically true as
*

*> we have seen above.
*

*> If w is not a terminal world, then there are worlds x, z, y ...
*

*> with wRx, wRz, wRy, ...
*

*> The fact that ([](p->q) & []p) is true at w entails that
*

*> both [](p->q) and []p are true at w. But this, by Kripke
*

*> semantics, entails that both (p->q) and p are true in
*

*> the accessible worlds x, z, y, ..., and so, q is true
*

*> in all those worlds (all world obeys classical logics and are
*

*> thus closed for modus ponens). But if q is true in all
*

*> accessible worlds from w, then, by Kripke again, []q must be
*

*> true in w.
*

*>
*

*> * * *
*

*>
*

*> In fact some other frames will play an important role, so let us
*

*> introduce them now. (The definition which follows are somehow
*

*> less standart).
*

*>
*

*> I will say, by definition, that a world is transitory iff it is
*

*> not terminal.
*

*>
*

*> A frame (W,R) will be said ideal if all worlds in W are transitory.
*

*> (there is simply no cul-de-sac worlds in an ideal frame).
*

*>
*

*> A frame will be said realist if there is a terminal world accessible
*

*> from any transitory world of that frame.
*

*> (You can "die" at every world in a realist frame, as you are
*

*> "immortal" in an ideal frame).
*

*>
*

*> The corresponding theorems (exercice) are
*

*>
*

*> 5. (W,R) respects []p -> <>p iff (W,R) is ideal,
*

*>
*

*> 6. (W,R) respects <>p -> -[]<>p iff (W,R) is realist,
*

*>
*

*> Do you have recognize <>p -> -[]<>p ?
*

*> Yes. It is our modal form of Godel theorem (with TRUE instead of p).
*

*> More will be said.
*

*>
*

*> My next post will anticipate toward the end of my proof to glimpse
*

*> the "quasi-appearance of Hilbert Space" when we explain the UDA TE
*

*> to the guardian angel of the sound machine.
*

*>
*

*> And, just because you promise me a prize for deriving SE from
*

*> the "psychology of machine" I tell you that I have decided
*

*> to call the modal formula (the one for the symmetrical frame):
*

*>
*

*> p->[]<>p,
*

*>
*

*> the little abstract Schroedinger Equation (LASE),
*

In English:

If p is true in one world then in the set of all worlds, there is at least on

world in which p is true....OK, isn't that obvious? But to connect this to SE,

where is the uncertainty? And in which world is the observer located?

*>
*

*>
*

*> as I have called before the (godel-like) formula
*

*>
*

*> <>p -> -[]<>p
*

*>
*

*> the first theorem of machine's psychology. (FTMP)
*

*>
*

*> And our goal is to find a natural bridge from FTMP to LASE.
*

*>
*

In English:

if there is at least one world where p is true, then it is false that in the set

of all worlds there is at least one world where p is true.

This statement seems inconsistent. Where does the choice between completeness

and consistency come into play?

*>
*

*> I must say that I have believe for a quite long time that this was
*

*> impossible. But then when you take the definition of knowledge
*

*> belief, observation/perception in Plato's thaetetus, then the
*

*> aritmetical translation of the UDA TE will lead us directly toward
*

*> the solution.
*

*> And so you will be obliged to give me the prize (at least
*

*> a little abstract price!).
*

*>
*

*> Infinite thanks for your patience. All of you.
*

*>
*

*> Bruno
*

*>
*

*> I hope your binder will not explode.
*

No the binder is fine.... my head however is getting swollen :-)

Again thanks for this course in logic

George

Received on Sun Apr 08 2001 - 13:58:35 PDT

Date: Sun, 08 Apr 2001 13:52:45 -0700

Marchal wrote:

but it is getting fatter....

I feel like Zeno... Or maybe like Sisyphus....:-)

Thanks

I am not sure if you gave a definition of reflexive, transitive and symmetric

I find it easier to comprehend these theorems if I state them in English:

1) If p is true in all worlds in set R, then p is true in world A. This is true

if A belongs to set R. (we could use a Venn diagram to prove this) ...... OK

2) if p is true in all worlds in set T then p is true in all worlds in set T in

all worlds in set T (????) There seems to be two levels of "in all worlds"

Slightly confusing

I don't get this one.... in English.

3) If p is true in world A then in all worlds in set R there is at least one

world B where p is true. This statement is true if A =B........ OK

4) If the relation "p implies q" is true in all worlds in set S, then, if p is

true in all world in S then q is true in all world in S... OK

In English:

If p is true in one world then in the set of all worlds, there is at least on

world in which p is true....OK, isn't that obvious? But to connect this to SE,

where is the uncertainty? And in which world is the observer located?

In English:

if there is at least one world where p is true, then it is false that in the set

of all worlds there is at least one world where p is true.

This statement seems inconsistent. Where does the choice between completeness

and consistency come into play?

No the binder is fine.... my head however is getting swollen :-)

Again thanks for this course in logic

George

Received on Sun Apr 08 2001 - 13:58:35 PDT

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